Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 3, Article 76, 2005
GEOMETRIC INEQUALITIES FOR A SIMPLEX
SHIGUO YANG
D EPARTMENT OF M ATHEMATICS
A NHUI I NSTITUTE OF E DUCATION
H EFEI , 230061, P.R. C HINA .
sxx@ahieedu.net.cn
Received 18 October, 2004; accepted 11 March, 2005
Communicated by J. Sándor
A BSTRACT. In this paper, we study a problem of geometric inequalities for an n-simplex. Some
new geometric inequalities for a simplex are established. As special cases, some known inequalities are deduced.
Key words and phrases: Simplex, Volume, Inradius, Circumradius, Inequality.
2000 Mathematics Subject Classification. 52A40, 51K16.
1. I NTRODUCTION
Let σn be an n-dimensional simplex in the n-dimensional Euclidean space E n ,
τ = {A0 , A1 , . . . , An } denote the vertex set of σn , V the volume of σn , R and r the circumradius and inradius of σn , respectively. For i = 0, 1, . . . , n, let ri be the radius of ith escribed
sphere of σn , Fi the area of the ith face fi = A0 · · · Ai−1 Ai+1 · · · An of σn . Let P be an arbitrary
interior point of the simplex σn , di the distance from the point P to the ith face fi of σn , hi the
altitude of σn from vertex Ai for i = 0, 1, . . . , n.
Let a0 , a1 and a2 denote the edge-lengths of triangle A0 A1 A2 (2-dimensional simplex). An
important inequality for a triangle was established by Janić (see [1]) as follows:
(1.1)
a20
a2
a2
+ 1 + 2 ≥ 4.
r 1 r2 r2 r 0 r0 r 1
Let P be an arbitrary interior point of the triangle A0 A1 A2 . Gerasimov (see [2]) obtained an
inequality for the triangle A0 A1 A2 as follows:
(1.2)
d1 d2 d2 d0 d0 d1
1
+
+
≤ .
a1 a2 a2 a0 a0 a1
4
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
The author would like to express his thanks to the editor for his kind help and invaluable suggestions in the formatting and writing of this
paper.
096-04
2
S HIGUO YANG
2. M AIN R ESULTS
We will extend inequalities (1.1) and (1.2) to an n-dimensional simplex. Our main results are
contained in the following theorem:
Theorem 2.1. For the n-dimensional simplex σn we have
2
n
n/(n−1)
X
Fi
(n − 1)n n3n /2(n−1)
(2.1)
≥ n
,
(n−2)/2 (n!)n/(n−1)
r
·
·
·
r
r
·
·
·
r
n
(n
+
1)
0
i−1
i+1
n
i=0
with equality iff the simplex σn is regular.
By letting n = 2 in relation (2.1), inequality (1.1) is reobtained.
Theorem 2.2. Let P be an arbitrary interior point of the simplex σn , and let θ ∈ (0, 1] be a real
number. Then we have
n
X
d0 · · · di−1 di+1 · · · dn
(n!)2θ
(2.2)
≤
V n−2(n−1)θ ,
2θ−1
(n−1)(1−θ)
n(3θ−1)
(F0 · · · Fi−1 Fi+1 · · · Fn )
(n + 1)
n
i=0
with equality iff the simplex σn is regular and the point P is the circumcenter of σn .
If we take θ =
n
2(n−1)
in inequality (2.2), we obtain the following corollary:
Corollary 2.3. Let P be an arbitrary interior point of the simplex σn . Then we have
n
X
(n!)n/(n−1)
d0 · · · di−1 di+1 · · · dn
≤
,
(2.3)
1/(n−1)
(n−2)/2 nn(n+2)/2(n−1)
(F
·
·
·
F
F
·
·
·
F
)
(n
+
1)
0
i−1
i+1
n
i=0
with equality iff the simplex σn is regular and the point P is the circumcenter of σn .
If n = 2 in inequality (2.3), then inequality (1.2) follows from inequality (2.3).
By taking θ = 21 in inequality (2.2), we obtain a generalization of Gerber’s inequality as
follows:
Corollary 2.4. Let P be arbitrary interior point of the simplex σn . Then
n
X
n!
(2.4)
d0 · · · di−1 di+1 · · · dn ≤
V,
(n + 1)(n−1)/2 nn/2
i=0
with equality iff the simplex σn is regular.
Using inequality (2.4) and the arithmetic-geometric mean inequality we get Gerber’s inequality [3] as follows:
n
Y
(n!)(n+1)/n
(2.5)
di ≤ (n+1)/2
V (n+1)/n .
1/2n
n
(n
+
1)
i=0
Theorem 2.5. Let P be an arbitrary interior point of the simplex σn . Then we have
n
X
1
r
(2.6)
≥ (n + 1)nn+1 · n+1 ,
d · · · di−1 di+1 · · · dn
R
i=0 0
with equality iff the simplex σn is regular and the point P is the circumcenter of σn .
If the point P is the incenter I of the simplex σn , i.e. di = r(i = 0, 1, . . . , n), then the
following n-dimensional Euler inequality stated in [4] is obtained from (2.6):
(2.7)
J. Inequal. Pure and Appl. Math., 6(3) Art. 76, 2005
R ≥ nr.
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G EOMETRIC I NEQUALITIES FOR A S IMPLEX
3
3. L EMMAS AND P ROOFS OF T HEOREMS
To prove the theorems stated above, we need some lemmas as follows.
Let mi (i = 0, 1, . . . , n) be positive numbers, Vi0 i1 ···ik denote the k-dimensional volume of
the k-dimensional simplex Ai0 Ai1 · · · Aik for Ai0 , Ai1 , . . . , Aik ∈ τ . Put
X
mi0 mi1 · · · mik Vi20 i1 ···ik ,
Mk =
(1 ≤ k ≤ n),
M0 =
0≤i0 <i1 <···<ik ≤n
n
X
mi .
i=0
Lemma 3.1. For positive numbers mi (i = 0, 1, . . . , n) and the n-dimensional simplex σn , we
have
(3.1)
Mkl ≥
[(n − l)!(l!)3 ]k
(n! · M0 )l−k Mlk ,
[(n − k)!(k!)3 ]l
(1 ≤ k < l ≤ n),
with equality iff the simplex σn is regular and m0 = m1 = · · · = mn .
Lemma 3.2.
(3.2)
n
Y
!
n
n2 −1
Fi
i=0
1
≥
(n + 1)1/2
n3n
n!2
1
2(n−1)
V n,
with equality iff the simplex σn is regular.
For the proof of Lemmas 3.1 and 3.2, the reader is referred to [5] or [1].
Lemma 3.3.
(3.3)
n
X
h0 · · · hi−1 hi+1 · · · hn
i=0
r0 · · · ri−1 ri+1 · · · rn
≥ (n + 1)(n1 )n ,
with equality iff the simplex σn is regular.
For the proof of Lemma 3.3, see [5].
Lemma 3.4.
nn/2 (n + 1)(n+1)/2 n
r ,
n!
with equality iff the simplex σn is regular.
V ≥
(3.4)
This is also known, see [5] or [1].
Proof of Theorem 2.1. Without loss of generality, let F0 ≤ F1 ≤ · · · ≤ Fn . By the known
formula ([1])
(3.5)
nV
,
j=0 Fj − 2Fi
ri = Pn
(i = 0, 1, . . . , n),
it follows that r0 ≤ r1 ≤ · · · ≤ rn and
Q
1
1
1
≤ Q
≤ ··· ≤ Q
.
Fj rj
Fj rj
Fj rj
j=1
J. Inequal. Pure and Appl. Math., 6(3) Art. 76, 2005
j=0
j6=i
j=0
j6=n
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4
S HIGUO YANG
Using the Chebyshev inequality, we have
! n
n
n
n/(n−1)
Y
X F 1/(n−1)
X
Fi
i
Q
Q
(3.6)
=
Fi
Fj rj
r
j
i=0
i=0
i=0
j=0
j6=i
j=0
j6=i

1
≥
n+1
n
Y
!
Fi
i=0
n
X
i=0
!
1/(n−1)
Fi

n
X
1 

.
Q


F
r
j j
i=0
j=0
j6=i
Substituting Fj = nV
(j = 0, 1, . . . , n) into the right side of inequality (3.6) and using the
hj
arithmetic-geometric mean inequality we get
! n
!
n
n
n
n/(n−1)
X
Y
X 1/(n−1)
Fi
1
1 X h0 · · · hi−1 hi+1 · · · hn
Q
(3.7)
≥
Fi
Fi
·
rj
n + 1 i=0
(nV )n i=0 r0 · · · ri−1 ri+1 · · · rn
i=0
i=0
j=0
j6=i
≥
n
Q
n2
(n2 −1)
Fi
i=0
(nV )n
n
X
h0 · · · hi−1 hi+1 · · · hn
i=0
r0 · · · ri−1 ri+1 · · · rn
.
By inequalities (3.7), (3.2) and (3.3) we obtain relation (2.1). It is easy to see that equality in
(2.1) holds iff the simplex σn is regular. The proof of Theorem 2.1 is thus complete.
Proof of Theorem 2.2. Taking k = n − 1, l = n in inequality (3.1), we can write
!n
! n
!n−1
n
n
X
Y
n3n X
2
mi
mi
V 2(n−1) .
(3.8)
m0 · · · mi−1 mi+1 · · · mn Fi
≥ 2
n!
i=0
i=0
i=0
By putting m0 · · · mi−1 mi+1 · · · mn = λi Fi−2 (i = 0, 1, . . . , n) in equality (3.8), we get
!n n
!
! n
!
n
n
2(n−1)
Y
Y
X F2
1X
(nV
)
i
(3.9)
λi
Fi2 ≥
λi
.
2
n i=0
(n
−
1)!
λ
i
i=0
i=0
i=0
We now prove that the following inequality (3.10) is valid for any number θ ∈ (0, 1]:
!n n
!n
!
n
n
n
2θ
Y
X
X
1X
F
(n + 1)2(n−1)θ (nV )2(n−1)θ
i
(3.10)
λi
Fi2θ ≥
λi
·
.
n(1−θ)
2θ
n i=0
λ
n
(n
−
1)!
i
i=0
i=0
i=0
When θ = 1, inequalities (3.10) and (3.9) are the same, so inequality (3.10) is valid for θ = 1.
For θ ∈ (0, 1), using inequality (3.9) we have
!n n
n
Y
1X
(3.11)
λi
Fi2θ
n i=0
i=0
"
!n n
#θ "
!n #1−θ
n
n
Y
X
1X
1
=
λi
Fi2 ·
λi
n i=0
n
i=0
i=0
"
! n
!#θ "
!n #1−θ
n
n
X F2
X
(nV )2(n−1) Y
1
i
≥
λi
·
λi
.
(n − 1)!2
λ
n i=0
i
i=0
i=0
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G EOMETRIC I NEQUALITIES FOR A S IMPLEX
5
By Maclaurin’s inequality ([1]) we have
n
1 X
λ0 · · · λi−1 λi+1 · · · λn
n + 1 i=0
! n1
n
1 X
≤
λi ,
n + 1 i=0
i.e.
n
(3.12)
1X
λi
n i=0
!n
(n + 1)n−1
≥
nn
n
Y
!
λi
i=0
n
X
1
λ
i=0 i
!
.
From (3.11) and (3.12) we can write
#1−θ
" n !n n
!" n 1 #θ
1
1−θ
n
n
Y
X
X F 2θ θ
X
1X
1
i
(3.13)
·
λi
Fi2θ ≥
λi
θ
n i=0
λ
λ1−θ
i
i
i=0
i=0
i=0
i=0
θ
1−θ (n + 1)n−1
(nV )2(n−1)
×
.
nn
(n − 1)!2
By Hölder’s inequality ([1]) we have
" n " n #1−θ
1 #θ
1
1−θ
n
X F 2θ θ
X
X
1
Fi2θ
i
(3.14)
·
≥
.
λi
λθi
λ1−θ
i
i=0
i=0
i=0
Using (3.13) and (3.14) we get relation (3.9).
P
Taking λi = di Fi (i = 0, 1, . . . , n) in equality (3.9) and noting the fact that ni=0 di Fi = nV ,
we get inequality (2.2). It is easy to prove that equality in (2.2) holds iff the simplex σn is regular
and the point P is the circumcenter of σn . The proof of Theorem 2.2 is thus complete.
Proof of Theorem 2.5. Inequality (3.9) can be written also as
n
(3.15)
n3n 2(n−1) X
V
λ0 · · · λi−1 λi+1 · · · λn Fi2 ≤
2
n!
i=0
0
σn0
n
X
!n
λi
i=0
n
Y
Fi2 .
i=0
Let V denote the volume of the n-dimensional simplex
=
· · · A0n , Fi0 being the area
of the ith face fi0 of σn0 . By Cauchy’s inequality and inequality (3.15), we have
(3.16)
A00 A01
n
n3n n−1 0 n−1 X
V
(V
)
λ0 · · · λi−1 λi+1 · · · λn Fi Fi0
2
n!
i=0
"
# 12
n
3n
X
n
≤
V 2(n−1)
λ0 · · · λi−1 λi+1 · · · λn Fi2
2
n!
i=0
"
# 12
n
3n
X
n
2(n−1)
2
×
(V 0 )
λ0 · · · λi−1 λi+1 · · · λn (Fi0 )
2
n!
i=0
!
! n
!
n
n
n
X
Y
Y
≤
λi
Fi
Fi0 .
i=0
i=0
i=0
If we suppose that σn0 is a regular simplex with F00 = F10 = · · · = Fn0 = 1. then
1
2 2(n−1)
n!
V 0 = (n + 1)1/2
,
n3n
J. Inequal. Pure and Appl. Math., 6(3) Art. 76, 2005
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6
S HIGUO YANG
so inequality (3.16) becomes
n
(n + 1)(n−1)/2 n3n/2 n−1 X
V
λ0 · · · λi−1 λi+1 · · · λn Fi ≤
n!
i=0
(3.17)
n
X
!n
λi
i=0
n
Y
Fi .
i=0
By letting λ0 = λ1 = · · · = λn = 1 in inequality (3.17), we get
1
n3n/2(n−1)
≥ 1/(n−1)
V
n!
(n + 1)(n+1)/2(n−1)
(3.18)
n
X
i=0
1
F0 · · · Fi−1 Fi+1 · · · Fn
1
! (n−1)
.
Now by Cauchy’s inequality we have
n
X
!
d0 · · · di−1 di+1 · · · dn
i=0
n
X
i=0
1
d0 · · · di−1 di+1 · · · dn
!
≥ (n + 1)2 ,
i.e.
n
X
(3.19)
i=0
1
≥ P
n
d0 · · · di−1 di+1 · · · dn
(n + 1)2
.
d0 · · · di−1 di+1 · · · dn
i=0
Using (3.19), (2.4) and (3.18), we get
n
X
1
(3.20)
d · · · di−1 di+1 · · · dn
i=0 0
≥
(n + 1)(n+3)/2 nn/2 1
·
n!
V
1
! (n−1)
1
(n−1)
2
2
n
(n + 1)(n +n−4)/2(n−1) nn /2(n−1) X
1
Qn
≥
Fi
.
(n − 1)!n/(n−1)
F
i
i=0
i=0
Pn
nV
By inequality (3.20), formula i=0 Fi = r and the known inequality ([1]):
n
Y
2
(n + 1)(n −1)/2
2
Fi ≤ n+1 (n2 −3n−4)/2 Rn −1 ,
n! n
i=0
(3.21)
we get
(3.22)
n
X
i=0
1
d0 · · · di−1 di+1 · · · dn
1
(n−1)
V
1
≥ (n + 1)
n
· n!
· n+1 .
r
R
Relations (3.22) and (3.4) imply inequality (2.6). It is easy to prove that equality in (2.6) holds
iff the simplex σn is regular and the point P is the circumcenter of σn . The proof of Theorem
2.5 is thus complete.
(n−3)/2(n−1) (2n2 −n−2)/2(n−1)
1/(n−1)
R EFERENCES
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[2] Ju.I. GERASIMOV, Problem 848, Mat. v Škole, 4 (1971), 86.
[3] L. GERBER, The orthecentric simplex as an extreme simplex, Pacific. J. Math., 56 (1975), 97–111.
J. Inequal. Pure and Appl. Math., 6(3) Art. 76, 2005
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G EOMETRIC I NEQUALITIES FOR A S IMPLEX
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[4] M.S. KLAMKIN, Problem 85–26, SIAM. Rev., 27(4) (1985), 576.
[5] J.Zh. ZHANG AND L. YANG, A class of geometric inequalities concerning the masspoint system,
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J. Inequal. Pure and Appl. Math., 6(3) Art. 76, 2005
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